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----------- Graph Isomorphism
An example might go a long way to explain this concept; this one comes from graph theory [619,622]. A graph is a network of lines connecting different points. If two graphs are identical except for the names of the points, they are called isomorphic. For an extremely large graph, finding whether two graphs are isomorphic can take centuries of computer time; it’s one of those NP-complete problems discussed in Section 11.1. Assume that Peggy knows the isomorphism between the two graphs, G1 and G2. The following protocol will convince Victor of Peggy’s knowledge: (1) Peggy randomly permutes G1 to produce another graph, H, that is isomorphic to G1. Because Peggy knows the isomorphism between H and G1, she also knows the isomorphism between H and G2. For anyone else, finding an isomorphism between G1 and H or between G2 and H is just as hard as finding an isomorphism between G1 and G2. (2) Peggy sends H to Victor. (3) Victor asks Peggy either to: (a) prove that H and G1 are isomorphic, or (b) prove that H and G2 are isomorphic. (4) Peggy complies. She either: (a) proves that H and G1 are isomorphic, without proving that H and G2 are isomorphic, or (b) proves that H and G2 are isomorphic, without proving that H and G1 are isomorphic. (5) Peggy and Victor repeat steps (1) through (4) n times. If Peggy does not know an isomorphism between G1 and G2, she cannot create graph H which is isomorphic to both. She can create a graph that is either isomorphic to G1 or one that is isomorphic to G2. Like the previous example, she has only a 50 percent chance of guessing which proof Victor will ask her to perform in step (3). This protocol doesn’t give Victor any useful information to aid him in figuring out an isomorphism between G1 and G2. Because Peggy generates a new graph H for each round of the protocol, he can get no information no matter how many rounds they go through the protocol. He won’t be able to figure out an isomorphism between G1 and G2 from Peggy’s answers. In each round, Victor receives a new random permutation of H, along with an is...
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- Fall '10